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TAMU GENE 412 - Chi-Squared for Genetics

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GENE 412 1st Edition Lecture 3Outline of Last Lecture I. Hardy-Weinburg for multiple allelesII. Find allele frequency from genotype frequencyIII. Chi-Squared for Genetics ProblemsOutline of Current Lecture I. X2 Chi-Squared TestII. P <.05 Concludes?Current LectureI. X2 Chi-Squared Testa. Ho: population is in H.W. equilibriumb. Ha: Population is not in H.W. equilibriumc. Equationi. Summation of (O-E)2/(E) where O=observed number in population & E=Expected number in populationAccept HoReject HoHo True Correct α = riskHa True Beta error(β) 1-β =powerd. ExampleAA Aa aa40 40 20p22pq q2Step 1 Find Allele FrequenciesFrequency of A = (40+20)/100 = .6Frequency of a = 1-p = .4Step 2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Find Chi-Squared Value for each genotype and then add togetherp22pq q2Expected 36 48 16O-E 4 -8 4(O-E)216 64 16(O-E)2/E 16/36 64/48 16/1616/36+64/48+16/16=2.77Step 3 Calculate Degrees of Freedom (DF)DF = (# of genotypes - # of alleles)Step 4 If Chi-Square value is above set risk or Alpha Error and for the degrees of freedom for experiment then the population is not in Hardy-Weinburg(H.W.) and therefore one of the assumptions of H.W. is not met (see lecture 1 for H.W. assumptions)II. Other Test topics for Chi-Squared a. Risk factors assume the type of risk you want to take b. In genetics usually use p<.05 is an acceptable risk to reject Hoc. Reasons H.W. is rejected i. Two genes in perfect H.W. while other is notii. Mate selected based on gene or gene is lethaliii. Hard to score H.W.d. Tips for Chi-Squaredi. Never reject H.W. because it is a law; the population just may not be in H.W. for a particular genee. Vocabulary i. Endoreduplicated is 2N amount of DNAii. Inversions- Create inversion loops and that is their only way to pair1. Inversions In Drosophila heterozygotes inhibit the ability to

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